chapter9

# chapter9 - Problem 9.1 This is a variation on the example...

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Problem 9.1 This is a variation on the example on pages 300-301. n ε 2 n ε 1 g ε 2 g ε 1 exp ε 2 ε 1 kT . . We are given that n( ε .2)/n( ε .1)=1/1000 and we want to solve for T. Also, we know that g( ε .1)=2 and g( ε .2)=8. We want to solve 1 1000 8 2 exp ε 2 ε 1 . . for T. I'll set this up more generally. ε 2 ε 1 . ln n ε 2 n ε 1 g ε 1 g ε 2 . n ε 2 n ε 1 g ε 1 g ε 2 . exp ε 2 ε 1 . . ε 1 ε 2 ln n ε 2 n ε 1 g ε 1 g ε 2 . T ε 1 ε 2 kln n ε 2 n ε 1 g ε 1 g ε 2 . . Now substitute values and get the answer. ε 1 13.6 ε 2 13.6 4 n 1 1000 n 2 1 g 1 2 g 2 8 k 8.617 10 5 . since we are working with eV ε 's T ε 1 ε 2 n 2 n 1 g 1 g 2 . . T 1.427 10 4 . = ____________________________________________________________________________ Problem 9.2 This is another variation on the example on pages 300-301. n ε 2 n ε 1 g ε 2 g ε 1 exp ε 2 ε 1 . . We are given T=5000 and we want to solve for n( ε .2)/n( ε .1), etc. n14 .. ε n () 13.6 n 2 gn () 2n 2 . k 8.617 10 5 . since we are working with eV ε 's T 5000 1

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ratio m n , () gm gn exp ε m ε n kT . . ratio 1 1 , 1 = (just checking) ratio 2 1 , ( ) 2.092 10 10 . = ratio 3 1 , ( ) 5.871 10 12 . = ratio 4 1 , ( ) 2.25 10 12 . = ____________________________________________________________________________ Problem 9.7 vbar 13 2 vbar 2 = m/s vrms 1 2 3 2 2 vrms 2.236 = m/s ____________________________________________________________________________ Problem 9.8 T 273 20 k 8.62 10 5 . Boltzmann's constant in units of eV/K The average energy per molecule is E 3 2 k . T . E 0.038 = eV This is much less than the 10.2 eV needed to raise a hydrogen atom from its ground state to its first excited state: E 10.2 3.71421 10 3 . = ____________________________________________________________________________ Problem 9.9 k 8.62 10 5 . Boltzmann's constant in units of eV/K E 13.6 Binding energy of hydrogen in eV We need to solve E 3 2 k . T . for T. T 2E . 3k . T 1.052 10 5 . = or about 10500 K ____________________________________________________________________________ 2
Problem 9.10 T 20 273 h 6.63 10 34 . k 1.38 10 23 . Boltzmann's constant in mks units m2 1 6 . 1.66 . 10 27 . mass of oxygen E 3 2 k . T . average oxygen molecule energy p2 m . E . oxygen momentum λ h p λ 2.612 10 11 . = de Broglie wavelength λ 410 10 . 0.065 = so the de Broglie wavelength is only about 6.5 percent of the molecular diameter ____________________________________________________________________________ Problem 9.14 The flux is the number of neutrons per square meter per second. If we divide the flux by the average neutron velocity, we get the number of neutrons per cubic meter at any time in the beam port: neutrons volume neutrons m 2 s .

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## This note was uploaded on 02/16/2010 for the course CHEM 550 taught by Professor Abra during the Fall '09 term at Penn State.

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chapter9 - Problem 9.1 This is a variation on the example...

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